Physics Question? Two Object?

Question

Two stones are dropped from the edge of a 60meter cliff, the second stone 1.5 s after the first. How far below the cliff is teh second stone when the separation between the two stones is 36meters.

Answer 1

write equations of motion for the two objects, in general:

y(t)=v0t+1/2at^2 where v0 is initial speed (zero here), a=9.8 m/s/s and t=time

the trick here is to recognize that the first object is in the air 1.5 seconds longer than the second one, so we can write:

y(first)=4.9(t+1.5)^2

y(second)=4.9t^2

remember, here t represents the time the second rock is in the air, and t+1.5 is the time the first rock is in the air

when the difference between them is 36 m, we have:

y(first)-y(second)=36=4.9(t+1.5)^2-4.9...

expanding:

4.9(t^2+3t+2.25)-4.9t^2=36

14.7t+11=36

t=1.7s

when the second rock is in the air for 1.7s, the first rock is in the air for 3.2 s and there is at a position of

4.9(3.2)^2=50.2m below the clilff

(as a check, when the second rock is in the air for 1.7s it is 14.2 m below the cliff)